QM Inverse Problem: Finding V(x)

  • Thread starter hilbert2
  • Start date
  • Tags
    Inverse Qm
In summary, the conversation is about finding the potential energy function V(x) that results in a particular eigenvalue spectrum for a one dimensional single particle system. The eigenvalues are given by E_n = k n^(3/2), where k is a constant and n = 1,2,3,... Various approaches to finding V(x) are discussed, including using a unitary transformation, testing different potentials, and using the Weyl formula or WKB quantization condition. However, it is noted that the energy spectrum does not uniquely determine V(x) and some conceivable spectra may not correspond to any physical potential. The conversation concludes with a suggestion to try V(x) = g|x|^n and solve for the value of n using
  • #1
hilbert2
Science Advisor
Insights Author
Gold Member
1,595
605
Let's say we have a one dimensional single particle system that is described by a SE

-[itex]\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi(x)+V(x)\psi(x)=E\psi(x)[/itex]

We do not know what the potential energy function V(x) is, but we know that the eigenvalues
spectrum is

[itex]E_{n}=kn^{\frac{3}{2}}[/itex]

where k is a constant with dimensions of energy and n=1,2,3,...

How to find out what is the function V(x) that results in this energy spectrum? If the exponent
were 1 instead of 3/2 then it would be a harmonic oscillator potential and if the exponent were
2 it would be the "particle in box" potential. Do all possible exponents result in a V(x) that can be
written with elementary functions?

To find V(x), we should obviously

1. Write the Hamiltonian matrix in the basis where its diagonal (we can do that)

2. Unitary transform to position representation basis

3. Subtract the kinetic energy operator

But how do we find the right unitary transformation to do this?
 
Last edited:
Physics news on Phys.org
  • #2
I would test V(x)=|x| and if that does not work, the more general V(x)=|x|a with real a, but that is just a guess.
In the general case: No idea. I know that in scattering theory, a transformation of the spatial potential to a "momentum potential" helps, but I don't know if that can be used here.
 
  • #3
^ I don't right now have at hand a numerical code that solves for the eigenstates and eigenvalues in 1D potential, but if I have time to make one I'll try what happens with the V(x) = |x| potential.

One has to note that the energy spectrum does not uniquely determine V(x). For example, the energy levels are same in the harmonic potential V(x) = kx2 and the translated harmonic potential V(x) = k(x+c)2 (c is a constant). Also applying a parity transformation to V(x) does not change the energy levels.

Also, its obvious that some conceivable spectra don't correspond to any physical potential. Suppose the energy spectrum is bounded from upward and En converges to some maximum energy when n grows without bound. That is not physically possible.
 
  • #4
I did some searching with Google and found a lot of mathematical literature about this exact problem (reconstructing a potential from spectral or scattering data). Look like the mathematicians Gelfand, Levitan and Marchenko solved this problem in 1960's.

I didn't understand much of the literature as it was for mathematicians, and it didn't provide any simple recipe for finding V(x) when En is known...
 
  • #5
hilbert2 said:
Also, its obvious that some conceivable spectra don't correspond to any physical potential. Suppose the energy spectrum is bounded from upward and En converges to some maximum energy when n grows without bound. That is not physically possible.
That's exactly how the energy states for a 1/r2 potential (in atoms, for example) look like.
 
  • #6
mfb said:
That's exactly how the energy states for a 1/r2 potential (in atoms, for example) look like.

That's only the bound states. The scattering states where the electron's energy is higher than the ionization energy of hydrogen (for example), don't have an energy upper limit. Scattering states form the continuum part of the hydrogenic energy spectrum.
 
  • #7
That's only the bound states.
Sure, but if you include unbound states you never get an upper limit on the energy.

I would not call this "conceivable spectra" ;).
The same goes for spectra without a lower bound.
 
  • #8
mfb said:
The same goes for spectra without a lower bound.

I vaguely remember from a QFT course that the fact that Dirac field operators must anticommute, is proven by arguing that otherwise the field modes could have arbitrarily large negative energy...

Of course if the energy levels of some system went towards negative infinity, you could extract arbitrarily large amounts of usable energy from the system by forcing it into lower and lower energy levels.
 
  • #9
You can get an approximate answer via the semiclassical Weyl formula for the number of states N(E) below a given energy E:
[tex]N(E) = \int{dx \,dp\over 2\pi\hbar}\,\theta(E-H(x,p))[/tex]
where [itex]\theta(z)[/itex] is the unit step function, [itex]H(x,p)[/itex] is the classical hamiltonian, and [itex]x[/itex] and [itex]p[/itex] are both integrated from -∞ to +∞.

I suspect this is what is intended. An exact answer would be much too hard.
 
  • #10
Avodyne said:
You can get an approximate answer via the semiclassical Weyl formula for the number of states N(E) below a given energy E:
[tex]N(E) = \int{dx \,dp\over 2\pi\hbar}\,\theta(E-H(x,p))[/tex]
where [itex]\theta(z)[/itex] is the unit step function, [itex]H(x,p)[/itex] is the classical hamiltonian, and [itex]x[/itex] and [itex]p[/itex] are both integrated from -∞ to +∞.

I suspect this is what is intended. An exact answer would be much too hard.

In the problem, I already know what the function N(E) is, and I want to find the classical hamiltonian. I would have to somehow invert that formula.
 
  • #11
Try V(x) = g|x|^n, and find the value of n that gives the N(E) that you want.
 
  • #12
Actually it can be obtained in an approximate way by wkb quantization condition.If you choose a potential v(x)=∞ for x<0 and v(x)=mgx for x>0 and solve it using wkb quantisation condition(by taking into account the infinite discontinuity otherwise it will get wrong).This potential can be solved exactly using Airy functions but nonetheless an approximate solution(using wkb) will give a dependence like [n+(1/4)]3/2 for the eigenvalue.
 

1. What is the QM inverse problem?

The QM inverse problem refers to the challenge of determining the potential function (V(x)) of a quantum mechanical system based on its known wave function. This problem is important in understanding the behavior and properties of quantum systems.

2. How is the QM inverse problem solved?

The QM inverse problem is typically solved using mathematical techniques such as perturbation theory, variational methods, and numerical algorithms. These methods involve solving the Schrödinger equation and manipulating it to find the potential function that best fits the given wave function.

3. Why is the QM inverse problem important?

The QM inverse problem is important because it allows us to gain a deeper understanding of quantum systems and their underlying potential functions. It also has practical applications in various fields such as material science, chemistry, and physics.

4. What are some challenges in solving the QM inverse problem?

One of the main challenges in solving the QM inverse problem is the non-uniqueness of solutions. This means that there can be multiple potential functions that yield the same wave function, making it difficult to determine the exact potential. Additionally, the complexity of the Schrödinger equation and the need for accurate experimental data can also pose challenges.

5. Can the QM inverse problem be solved for all quantum systems?

No, the QM inverse problem cannot be solved for all quantum systems. It is only possible to solve it for simple systems with known wave functions, such as the hydrogen atom. For more complex systems, approximations and numerical methods are often used to find an approximate potential function.

Similar threads

  • Quantum Physics
Replies
19
Views
1K
  • Quantum Physics
Replies
2
Views
945
Replies
4
Views
1K
Replies
1
Views
633
Replies
17
Views
1K
  • Quantum Physics
Replies
17
Views
738
  • Quantum Physics
Replies
31
Views
2K
Replies
3
Views
756
Replies
4
Views
823
Replies
8
Views
2K
Back
Top